weighing 2 pounds attached to a spring stretches it 2 feet. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 0.5 ft/s. Determine the equation of motion. ************ Formulas Weight = mg=ks g=32ft/s2 w2 = (t)=ccos(wt)+c2sin(wt)

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A mass weighing 2 pounds attached to a spring stretches it 2 feet. The mass is initially released from a point 1 foot above the
equilibrium position with a downward velocity of 0.5 ft/s. Determine the equation of motion.
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Formulas
Weight = mg=ks
g=32ft/s2
w2 =
(t)=c1cos(wt)+c2sin(wt)
Transcribed Image Text:A mass weighing 2 pounds attached to a spring stretches it 2 feet. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 0.5 ft/s. Determine the equation of motion. *********** ************ Formulas Weight = mg=ks g=32ft/s2 w2 = (t)=c1cos(wt)+c2sin(wt)
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